integration problem

**katie** · May 31st 2008, 11:20 PM

Hi guys,

I'm super stuck with these two problems on one of my practice exams, can anyone help me out?

Find the integral between 4 and 3 of (u^2 + 1) / (u - 2)^2
and the antiderivative of cosxe^(sinx) dx

It asks to find the volume of the solid of revolution obtained by rotating, a full turn about the x-axis, the area between the x-axis and the curve y = 2 sin x, for x ∈ [ π , 3π ].

Thnanks

**flyingsquirrel** · June 1st 2008, 12:25 AM

Hi

Originally Posted by katie

I'm super stuck with these two problems on one of my practice exams, can anyone help me out?

Find the integral between 4 and 3 of (u^2 + 1) / (u - 2)^2

The hint is to rewrite the integral as follows (using $(u-2)=u^2-4u+4$

)

$ \begin{aligned} \int_3^4\frac{u^2+1}{(u-2)^2}\,\mathrm{d}u &= \int_3^4\frac{u^2-4u+4+4u-4+1}{(u-2)^2}\,\mathrm{d}u\\ &=\int_3^4\frac{u^2-4u+4}{(u-2)^2}\,\mathrm{d}u+ \int_3^4\frac{4u-3}{(u-2)^2}\,\mathrm{d}u\\ &=\int_3^4\mathrm{d}u+ \int_3^4\frac{4u-3}{(u-2)^2}\,\mathrm{d}u\\ \end{aligned} $

For the last integral, let's make appear the derivative of the denominator ( $2u-4$ ) at the numerator so that one can easily find an anti-derivative :

$\begin{aligned} \int_3^4\frac{4u-3}{(u-2)^2}\,\mathrm{d}u&=2\int_3^4\frac{2u-\frac{3}{2}}{(u-2)^2}\,\mathrm{d}u\\ &=2\int_3^4\frac{2u-4+4-\frac{3}{2}}{(u-2)^2}\,\mathrm{d}u\\ &=2\int_3^4\frac{2u-4}{(u-2)^2}\,\mathrm{d}u+2\int_3^4\frac{4-\frac{3}{2}}{(u-2)^2}\,\mathrm{d}u\\ &=2\int_3^4\frac{2u-4}{(u-2)^2}\,\mathrm{d}u+5\int_3^4\frac{1}{(u-2)^2}\,\mathrm{d}u\\ \end{aligned}$

If you don't see the what can be done with the last two integrals, substitute $x=u-2$ .

Hope that helps.

**mr fantastic** · June 1st 2008, 12:34 AM

Originally Posted by katie

Hi guys,

I'm super stuck with these two problems on one of my practice exams, can anyone help me out?

Find the integral between 4 and 3 of (u^2 + 1) / (u - 2)^2

Mr F says: ${\color{red} \frac{u^2 + 1}{u^2 - 4u + 4} = \frac{(u^2 - 4u + 4) + (4u - 3)}{u^2 - 4u + 4} = 1 + \frac{4u - 3}{u^2 - 4u + 4}}$ .

Now note that ${\color{red} \frac{4u - 3}{u^2 - 4u + 4} = \frac{(4u - 8) + 5}{u^2 - 4u + 4} = \frac{4u - 8}{u^2 - 4u + 4} + \frac{5}{u^2 - 4u + 4}}$ .

Therefore ${\color{red} \frac{u^2 + 1}{(u-2)^2} = 1 + \frac{5}{(u-2)^2} + \frac{4u - 8}{u^2 - 4u + 4}}$ .

See what you can do with that.

and the antiderivative of cosxe^(sinx) dx

Mr F says: Make the substitution ${\color{red} u = \sin x}$ .

It asks to find the volume of the solid of revolution obtained by rotating, a full turn about the x-axis, the area between the x-axis and the curve y = 2 sin x, for x ∈ [ π , 3π ].

Mr F says: By symmetry, it'll be twice the volume when the area from 0 to pi is rotated:

${\color{red} V = 2 \pi \int_{0}^{\pi} (2 \sin x)^2 \, dx = ......}$ .

Thnanks

..

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